Bernoulli's differential equation is an equation of the form
where n≠0,n≠1.
This equation can be rearranged using substitution
On practice differential equation Bernoulli usually does not lead to a linear equation, but is immediately solved using the same methods as a linear equation - either the Bernoulli method or the method of variation of an arbitrary constant.
Let's look at how to solve Bernoulli's differential equation using the substitution y=uv (Bernoulli's method). The solution scheme is the same as for .
Examples. Solve equations:
1) y’x+y=-xy².
This is Bernoulli's differential equation. Let's bring it to standard form. To do this, divide both parts by x: y’+y/x=-y². Here p(x)=1/x, q(x)=-1, n=2. But we don't need it to solve standard view. We will work with the recording form given in the condition.
1) Replacement y=uv, where u=u(x) and v=v(x) are some new functions of x. Then y’=(uv)’=u’v+v’u. We substitute the resulting expressions into the condition: (u’v+v’u)x+uv=-xu²v².
2) Let’s open the brackets: u’vx+v’ux+uv=-xu²v². Now let’s group the terms with v: v+v’ux=-xu²v² (I) (we don’t touch the term with degree v, which is on the right side of the equation). Now we require that the expression in brackets equals zero: u’x+u=0. And this is an equation with separable variables u and x. Having solved it, we will find u. We substitute u=du/dx and separate the variables: x·du/dx=-u. We multiply both sides of the equation by dx and divide by xu≠0:
(when finding u C we take it equal to zero).
3) In equation (I) we substitute =0 and the found function u=1/x. We have the equation: v’·(1/x)·x=-x·(1/x²)·v². After simplification: v’=-(1/x)·v². This is an equation with separable variables v and x. We replace v’=dv/dx and separate the variables: dv/dx=-(1/x)·v². We multiply both sides of the equation by dx and divide by v²≠0:
(we took -C so that, by multiplying both sides by -1, we could get rid of the minus). So, multiply by (-1):
(one could take not C, but ln│C│, and in this case it would be v=1/ln│Cx│).
2) 2y’+2y=xy².
Let's make sure that this is Bernoulli's equation. Dividing both parts by 2, we get y’+y=(x/2) y². Here p(x)=1, q(x)=x/2, n=2. We solve the equation using Bernoulli's method.
1) Replacement y=uv, y’=u’v+v’u. We substitute these expressions into the original condition: 2(u’v+v’u)+2uv=xu²v².
2) Open the brackets: 2u’v+2v’u+2uv=xu²v². Now let’s group the terms containing v: +2v’u=xu²v² (II). We require that the expression in brackets equals zero: 2u’+2u=0, hence u’+u=0. This is a separable equation for u and x. Let's solve it and find u. We substitute u’=du/dx, from where du/dx=-u. Multiplying both sides of the equation by dx and dividing by u≠0, we get: du/u=-dx. Let's integrate:
3) Substitute into (II) =0 and
Now we substitute v’=dv/dx and separate the variables:
Let's integrate:
The left side of the equality is a table integral, the integral on the right side is found using the integration by parts formula:
Substituting the found v and du using the integration by parts formula we have:
And since
Let's make C=-C:
4) Since y=uv, we substitute the found functions u and v:
3) Integrate the equation x²(x-1)y’-y²-x(x-2)y=0.
Let's divide both sides of the equation by x²(x-1)≠0 and move the term with y² to the right side:
This is Bernoulli's equation
1) Replacement y=uv, y’=u’v+v’u. As usual, we substitute these expressions into the original condition: x²(x-1)(u’v+v’u)-u²v²-x(x-2)uv=0.
2) Hence x²(x-1)u’v+x²(x-1)v’u-x(x-2)uv=u²v². We group the terms containing v (v² - do not touch):
v+x²(x-1)v’u=u²v² (III). Now we require that the expression in brackets be equal to zero: x²(x-1)u’-x(x-2)u=0, hence x²(x-1)u’=x(x-2)u. In the equation we separate the variables u and x, u’=du/dx: x²(x-1)du/dx=x(x-2)u. We multiply both sides of the equation by dx and divide by x²(x-1)u≠0:
On the left side of the equation is a tabular integral. Rational fraction on the right side you need to decompose into simple fractions:
At x=1: 1-2=A·0+B·1, whence B=-1.
At x=0: 0-2=A(0-1)+B·0, whence A=2.
ln│u│=2ln│x│-ln│x-1│. According to the properties of logarithms: ln│u│=ln│x²/(x-1)│, whence u=x²/(x-1).
3) In equality (III) we substitute =0 and u=x²/(x-1). We get: 0+x²(x-1)v’u=u²v²,
v’=dv/dx, substitute:
instead of C, we take - C, so that, by multiplying both sides by (-1), we get rid of the minuses:
Now let’s reduce the expressions on the right side to a common denominator and find v:
4) Since y=uv, substituting the found functions u and v, we get:
Self-test examples:
1) Let's make sure that this is Bernoulli's equation. Dividing both sides by x, we have:
1) Replacement y=uv, from where y’=u’v+v’u. We substitute these y and y’ into the original condition:
2) Group the terms with v:
Now we require that the expression in brackets equals zero and find u from this condition:
Let's integrate both sides of the equation:
3) In equation (*) we substitute =0 and u=1/x²:
Let's integrate both sides of the resulting equation.
An equation of the form y’ + P(x)y = Q(x), where P(x) and Q(x) are known functions of x, linear with respect to the function y and its derivative y’, is called a first-order linear differential equation.
If q(x)=0, the equation is called a linear homogeneous equation. q(x)=0 – linear inhomogeneous equation.
A linear equation is reduced to two equations with separable variables using the substitution y = u*v, where u = u(x) and v = v(x) are some auxiliary continuous functions.
So, y = u*v, y’ = u’*v + u * v’ (1),
then we rewrite the original equation in the form: u’*v + u * v’ + P(x)*v = Q(x) (2).
Since the unknown function y is sought as a product of two functions, one of them can be chosen arbitrarily, the other can be determined by equation (2).
Let us choose so that v’ + P(x)*v = 0 (3). For this, it is sufficient that v(x) be a partial solution of equation (3) (at C = 0). Let's find this solution:
V*P(x) ; = -;ln |v| = -;v = (4)
Substituting function (4) into equation (2), we obtain a second equation with separable variables, from which we find the function u(x):
u’ * = Q(x) ; du = Q(x) *; u = 
+C (5)
Finally we get:
y(x) = u(x)*v(x) = *( 
+C)
Bernoulli's equation:y’ + y = x* y 3
This equation has the form: y’ + P(x)*y = y’’ * Q(x), where P(x) and Q(x) are continuous functions.
If n = 0, then Bernoulli's equation becomes a linear differential equation. If n = 1, the equation becomes a separable equation.
In general, when n ≠ 0, 1, eq. Bernoulli is reduced to a linear differential equation using substitution: z = y 1- n
The new differential equation for the function z(x) has the form: z" + (1-n)P(x)z = (1-n)Q(x) and can be solved in the same ways as linear differentials .1st order equations.
20. Differential equations of higher orders.
Let's consider the equation that does not contain the function explicitly:
|
|
The order of this equation is reduced by one using substitution:
Indeed, then:
And we get an equation in which the order is lowered by one:
Diff. equations of order higher than the second have the form and , where are real numbers, and the function f(x) continuous on the integration interval X.
It is not always possible to solve such equations analytically and approximate methods are usually used. However, in some cases it is possible to find common decision.
Theorem.
General solution y 0
linear homogeneous differential equation on the interval X with continuous coefficients on X is a linear combination n linearly independent partial solutions of LODE 
with arbitrary constant coefficients 
, that is 
.
Theorem.
Common decision y linear inhomogeneous differential
equations on the interval X with continuous ones on the same
in between X coefficients and function f(x) represents the amount
Where y 0 is the general solution of the corresponding LODE, and is some particular solution of the original LODE.
Thus, the general solution of a linear inhomogeneous differential equation with constants
looking for coefficients in the form , where - some
his private solution, and 
– general solution of the corresponding homogeneous differential
equations
21. Tests and events. Types of events. Examples.
Testing is the creation of a certain set of conditions for the occurrence of events. Example: throwing a dice
Event – occurrence/non-occurrence of one or another test outcome; test result. Example: rolling the number 2
A random event is an event that may or may not occur during a given test. Example: rolling a number greater than 5
Reliable - an event that inevitably occurs during a given test. Example: rolling a number greater than or equal to 1
Possible - an event that can happen during a given test. Example: rolling the number 6
Impossible - an event that cannot occur during a given test. Example: rolling the number 7
Let A be some event. By an event opposite to it, we will understand an event consisting in the non-occurrence of event A. Designation: Ᾱ. Example: A – number 2 is rolled, Ᾱ – any other number is rolled
Events A and B are incompatible if the occurrence of one of them excludes the occurrence of the other in the same trial. Example: getting the numbers 1 and 3 on the same roll.
Events A and B are called joint if they can occur in one trial. Example: getting a number greater than 2 and the number 4 on the same roll.
22. Complete group of events. Examples.
A complete group of events - events A, B, C, D, ..., L, which are considered to be the only possible ones if, as a result of each test, at least one of them will definitely occur. Example: number 1, number 2, 3, 4, 5, 6 appear on the dice.
23. Event frequency. Statistical definition of probability.
Let n tests be carried out, and event A occurs m times. This m:n ratio is the frequency of occurrence of event A.
Def. The probability of a random event is a constant number associated with a given event, around which the frequency of occurrence of this event fluctuates in long series of tests.
Probability is calculated before the experiment, and frequency after it.
24. Classic definition of probability. Properties of event probability.
The probability of an event x is the ratio of the number of outcomes favorable to event A to the total number of all equally possible pairwise incompatible and uniquely possible outcomes of the experiment. P(A) =
Event probability properties:
For any event A 0<=m<=n
Dividing each term by n, we obtain for the probability of any event A: 0<=Р(А) <=1
If m=0, then the event is impossible: P(A)=0
If m=n, then the event is reliable: P(A)=1
If m 25. Geometric definition of probability. Examples. The classical definition of probability requires consideration of a finite number of elementary outcomes, and equally possible ones. But in practice there are often tests in which the number of possible outcomes is infinite. ODA. If a point randomly appears in a one-dimensional, two-dimensional or three-dimensional region of measure S (a measure is its length, area or volume), then the probability of its appearance in part of this region of measure S is equal to where S is a geometric measure expressing the total number all possible and equally possible outcomes of this trial, and S i – a measure expressing the number of outcomes favorable to event A. Example 1. A circle of radius R is placed in a smaller circle of radius r. Find the probability that a point thrown at random into the larger circle will also fall into the small circle. Example 2. Let a segment of length l be included in a segment of length L. Find the probability of event A “a randomly thrown point falls on a segment of length l.” Example 3 . A point is randomly selected in the circle. What is the probability that its distance to the center of the circle is greater than half? Example 4. The two persons agreed to meet at a certain place between two and three o'clock in the afternoon. The first person to arrive waits for the other person for 10 minutes and then leaves. What is the probability of these persons meeting if each of them can arrive at any time during the specified hour, regardless of the other? 26. Elements of combinatorics: Placement, permutation, combinations. 1) Permutation is called the order established in a finite set. The number of all different permutations is calculated by the formula 2) Placement n from elements by m called anything
a subset of the main set containing m elements. 3) Combination from n elements by elements by m disorderly
a subset of the main set containing elements. The differential equation y" +a 0 (x)y=b(x)y n is called Bernoulli's equation.
Purpose of the service. An online calculator can be used to check the solution Bernoulli differential equations. Example 1. Find the general solution to the equation y" + 2xy = 2xy 3. This is Bernoulli's equation for n=3. Dividing both sides of the equation by y 3 we get. Make a change. Then and therefore the equation is rewritten as -z" + 4xz = 4x. Solving this equation by the method of variation of an arbitrary constant, we obtain Example 2. y"+y+y 2 =0 Divide by y 2 We make a replacement: We get: -z" + z = -1 or z" - z = 1 Example 3. xy’+2y+x 5 y 3 e x =0 Bernoulli's equation is one of the most famous nonlinear differential equations of the first order. It is written in the form Where a(x) And b(x) are continuous functions. If elements by= 0, then Bernoulli's equation becomes a linear differential equation. In the case when elements by= 1, the equation becomes a separable equation. In general, when elements by≠ 0.1, Bernoulli's equation is reduced to a linear differential equation using the substitution New differential equation for the function z(x) has the form and can be solved using the methods described on the page First-order linear differential equations. BERNOULI METHOD. The equation under consideration can be solved by Bernoulli's method. To do this, we look for a solution to the original equation in the form of a product of two functions: where u, v- functions from x. Differentiate: Substitute into the original equation (1): First order differential equation of the form called equation in total differentials, if its left side represents the total differential of some function, i.e. Theorem. In order for equation (1) to be an equation in total differentials, it is necessary and sufficient that in some simply connected domain of change of variables the condition is satisfied The general integral of equation (1) has the form or – a measure expressing the number of outcomes favorable to event A.
Solve differential equation.
Solution.
Let's check that this equation is a total differential equation:
so that is condition (2) is satisfied. Thus, this equation is an equation in total differentials and
therefore, where is still an undefined function.
Integrating, we get . The partial derivative of the found function must be equal to, which gives from where so that Thus,.
General integral of the original differential equation.
When integrating some differential equations, the terms can be grouped in such a way that easily integrable combinations are obtained.
Linear differential operator and its properties. The set of functions having on the interval ( a
, b
) no less n
derivatives, forms a linear space. Consider the operator L
n
(y
), which displays the function y
(x
), having derivatives, into a function having k
- n
derivatives. A first-order linear differential equation is an equation that is linear with respect to an unknown function and its derivative. It looks like \frac(dy)(dx)+p(x)y=q(x), where p(x) and q(x) are given functions of x, continuous in the region in which equation (1) needs to be integrated. If q(x)\equiv0 , then equation (1) is called linear homogeneous. It is a separable equation and has a general solution Y=C\exp\!\left(-\int(p(x))\,dx\right)\!, The general solution to the inhomogeneous equation can be found method of variation of an arbitrary constant, which consists in the fact that the solution to equation (1) is sought in the form Y=C(x)\exp\!\left(-\int(p(x))\,dx\right), where C(x) is a new unknown function of x. Example 1. Solve the equation y"+2xy=2xe^(-x^2) . Solution. Let us apply the method of variation of a constant. Consider the homogeneous equation y"+2xy=0, corresponding to this inhomogeneous equation. This is an equation with separable variables. Its general solution has the form y=Ce^(-x^2) . We look for a general solution to the inhomogeneous equation in the form y=C(x)e^(-x^2), where C(x) is an unknown function of x. Substituting, we get C"(x)=2x, whence C(x)=x^2+C. So, the general solution of the inhomogeneous equation will be y=(x^2+C)e^(-x^2) , where C - constant of integration. Comment. It may turn out that the differential equation is linear in x as a function of y. The normal form of such an equation is \frac(dx)(dy)+r(y)x=\varphi(y). Example 2. Solve the equation \frac(dy)(dx)=\frac(1)(x\cos(y)+\sin2y). Solution. This equation is linear if we consider x as a function of y: \frac(dx)(dy)-x\cos(y)=\sin(2y). We use the method of variation of an arbitrary constant. First we solve the corresponding homogeneous equation \frac(dx)(dy)-x\cos(y)=0, which is an equation with separable variables. Its general solution has the form x=Ce^(\sin(y)),~C=\text(const). We look for a general solution to the equation in the form x=C(y)e^(\sin(y)), where C(y) is an unknown function of y. Substituting, we get C"(y)e^(\sin(y))=\sin2y or C"(y)=e^(-\sin(y))\sin2y. From here, integrating by parts, we have \begin(aligned)C(y)&=\int(e^(-\sin(y))\sin2y)\,dy=2\int(e^(-\sin(y))\cos(y) \sin(y))\,dy=2\int\sin(y)\,d(-e^(-\sin(y)))=\\ &=-2\sin(y)\,e^ (-\sin(y))+2\int(e^(-\sin(y))\cos(y))\,dy=C-2(\sin(y)+1)e^(-\ sin(y)),\end(aligned) So, C(y)=-2e^(-\sin(y))(1+\sin(y))+C. X=Ce^(\sin(y))-2(1+\sin(y)) The original equation can also be integrated as follows. We believe Y=u(x)v(x), where u(x) and v(x) are unknown functions of x, one of which, for example v(x), can be chosen arbitrarily. Substituting y=u(x)v(x) into , after transformation we get Vu"+(pv+v")u=q(x). Determining v(x) from the condition v"+pv=0, we then find from vu"+(pv+v")u=q(x) the function u(x) and, consequently, the solution y=uv of the equation \frac(dy)(dx)+p(x)y=q(x). As v(x) we can take any frequent solution of the equation v"+pv=0,~v\not\equiv0. Example 3. Solve the Cauchy problem: x(x-1)y"+y=x^2(2x-1),~y|_(x=2)=4. Solution. We are looking for a general solution to the equation in the form y=u(x)v(x) ; we have y"=u"v+uv". Substituting the expression for y and y" into the original equation, we will have X(x-1)(u"v+uv")+uv=x^2(2x-1) or x(x-1)vu"+u=x^2(2x-1) We find the function v=v(x) from the condition x(x-1)v"+v=0. Taking any particular solution of the last equation, for example v=\frac(x)(x-1) and substituting it, we get the equation u"=2x-1, from which we find the function u(x)=x^2-x+C. Therefore, the general solution to the equation x(x-1)y"+y=x^2(2x-1) will Y=uv=(x^2-x+C)\frac(x)(x-1), or y=\frac(Cx)(x-1)+x^2. Using the initial condition y|_(x=2)=4, we obtain the equation for finding C 4=\frac(2C)(2-1)+2^2, from where C=0 ; so the solution to the stated Cauchy problem will be the function y=x^2. Example 4. It is known that there is a relationship between the current i and the electromotive force E in a circuit having resistance R and self-inductance L E=Ri+L\frac(di)(dt), where R and L are constants. If we consider E a function of time t, we obtain a linear inhomogeneous equation for the current strength i: \frac(di)(dt)+\frac(R)(L)i(t)=\frac(E(t))(L). Find the current strength i(t) for the case when E=E_0=\text(const) and i(0)=I_0 . Solution. We have \frac(di)(dt)+\frac(R)(L)i(t)=\frac(E_0)(L),~i(0)=I_0. The general solution of this equation has the form i(t)=\frac(E_0)(R)+Ce^(-(R/L)t). Using the initial condition (13), we obtain from C=I_0-\frac(E_0)(R), so the desired solution will be I(t)=\frac(E_0)(R)+\left(I_0-\frac(E_0)(R)\right)\!e^(-(R/L)t). This shows that at t\to+\infty the current strength i(t) tends to a constant value \frac(E_0)(R) . Example 5. A family C_\alpha of integral curves of the linear inhomogeneous equation y"+p(x)y=q(x) is given. Show that the tangents at the corresponding points to the curves C_\alpha defined by the linear equation intersect at one point (Fig. 13). Solution. Consider the tangent to any curve C_\alpha at the point M(x,y). The equation of the tangent at the point M(x,y) has the form \eta-q(x)(\xi-x)=y, where \xi,\eta are the current coordinates of the tangent point. By definition, at the corresponding points x is constant and y is variable. Taking any two tangents to the lines C_\alpha at the corresponding points, for the coordinates of the point S of their intersection, we obtain \xi=x+\frac(1)(p(x)), \quad \eta=x+\frac(q(x))(p(x)). This shows that all tangents to the curves C_\alpha at the corresponding points ( x is fixed) intersect at the same point S\!\left(x+\frac(1)(p(x));\,x+\frac(q(x))(p(x))\right). Eliminating the argument x in the system, we obtain the equation of the locus of points S\colon f(\xi,\eta)=0. Example 6. Find the solution to the equation y"-y=\cos(x)-\sin(x), satisfying the condition: y is limited at y\to+\infty . Solution. The general solution to this equation is y=Ce^x+\sin(x) . Any solution to the equation obtained from the general solution for C\ne0 will be unbounded, since for x\to+\infty the function \sin(x) is bounded and e^x\to+\infty . It follows that this equation has a unique solution y=\sin(x) , bounded at x\to+\infty , which is obtained from the general solution at C=0 . Bernoulli's differential equation looks like \frac(dy)(dx)+p(x)y=q(x)y^n, where n\ne0;1 (for n=0 and n=1 this equation is linear). Using variable replacement z=\frac(1)(y^(n-1)) Bernoulli's equation is reduced to a linear equation and integrated as a linear one. Example 7. Solve Bernoulli's equation y"-xy=-xy^3. Solution. Divide both sides of the equation by y^3: \frac(y")(y^3)-\frac(x)(y^2)=-x Making a variable change \frac(1)(y^2)=z\Rightarrow-\frac(2y")(y^3)=z", where \frac(y")(y^3)=-\frac(z")(2). After substitution, the last equation turns into a linear equation -\frac(z")(2)-xz=-x or z"+2xz=2x, the general solution of which is z=1+Ce^(-x^2). \frac(1)(y^2)=1+Ce^(-x^2) or y^2(1+Ce^(-x^2))=1. Comment. Bernoulli's equation can also be integrated by the method of variation of a constant, like a linear equation, and using the substitution y(x)=u(x)v(x) . Example 8. Solve Bernoulli's equation xy"+y=y^2\ln(x). . Solution. Let us apply the method of variation of an arbitrary constant. The general solution of the corresponding homogeneous equation xy"+y=0 has the form y=\frac(C)(x). We look for the general solution of the equation in the form y=\frac(C(x))(x) , where C(x) - new unknown function. Substituting into the original equation, we have C"(x)=C^2(x)\frac(\ln(x))(x^2). To find the function C(x), we obtain an equation with separable variables, from which, by separating the variables and integrating, we find \frac(1)(C(x))=\frac(\ln(x))(x)+\frac(1)(x)+C~\Rightarrow~C(x)=\frac(x)( 1+Cx+\ln(x)). So, the general solution to the original equation y=\frac(1)(1+Cx+\ln(x)). Some first-order nonlinear equations can be reduced to linear equations or Bernoulli equations using a successfully found change of variables. Example 9. Solve the equation y"+\sin(y)+x\cos(y)+x=0. Solution. Let us write this equation in the form y"+2\sin\frac(y)(2)\cos\frac(y)(2)+2x\cos^2\frac(y)(2)=0.. Dividing both sides of the equation by 2\cos^2\frac(y)(2), we get \frac(y")(2\cos^2\dfrac(y)(2))+\operatorname(tg)\frac(y)(2)+x=0. Replacement \operatorname(tg)\frac(y)(2)=z\Rightarrow\frac(dz)(dx)=\frac(y")(\cos^2\dfrac(y)(2)) reduces this equation to linear \frac(dz)(dx)+z=-x, the general solution of which is z=1-x+Ce^(-x) . Replacing z by its expression in terms of y, we obtain the general integral of this equation \operatorname(tg)\frac(y)(2)=1-x+Ce^(-x). In some equations, the desired function y(x) may be under the integral sign. In these cases, it is sometimes possible to reduce this equation to a differential equation by differentiation. Example 10. Solve the equation x\int\limits_(x)^(0)y(t)\,dt=(x+1)\int\limits_(0)^(x)ty(t)\,dt,~x>0. Solution. Differentiating both sides of this equation with respect to x, we get \int\limits_(0)^(x)y(t)\,dt+xy(x)=\int\limits_(0)^(x)ty(t)\,dt+x(x+1)y (x) or Source of information
Since with n=0 a linear equation is obtained, and with n=1 - with separable variables, we assume that n ≠ 0 and n ≠ 1. Divide both sides of (1) by y n. Then, putting , we have . Substituting this expression, we get 
, or, which is the same thing, z" + (1-n)a 0 (x)z = (1-n)b(x). This is a linear equation that we know how to solve.
where 
or, what is the same, 
.
y"+y = -y 2
y"/y 2 + 1/y = -1
z=1/y n-1 , i.e. z = 1/y 2-1 = 1/y
z = 1/y
z"= -y"/y 2
Solution.
a) Solution through the Bernoulli equation.
Let's present it in the form: xy’+2y=-x 5 y 3 e x . This is Bernoulli's equation for n=3. Dividing both sides of the equation by y 3 we get: xy"/y 3 +2/y 2 =-x 5 e x. We make the replacement: z=1/y 2. Then z"=-2/y 3 and therefore the equation is rewritten in the form : -xz"/2+2z=-x 5 e x. This is a non-homogeneous equation. Consider the corresponding homogeneous equation: -xz"/2+2z=0
1. Solving it, we get: z"=4z/x
Integrating, we get:
ln(z) = 4ln(z)
z=x4. We now look for a solution to the original equation in the form: y(x) = C(x)x 4 , y"(x) = C(x)"x 4 + C(x)(x 4)"
-x/2(4C(x) x 3 +C(x)" x 4)+2y=-x 5 e x
-C(x)" x 5 /2 = -x 5 e x or C(x)" = 2e x . Integrating, we get: C(x) = ∫2e x dx = 2e x +C
From the condition y(x)=C(x)y, we obtain: y(x) = C(x)y = x 4 (C+2e x) or y = Cx 4 +2x 4 e x. Since z=1/y 2, we get: 1/y 2 = Cx 4 +2x 4 e x
(2) As v Let’s take any non-zero solution to the equation: 
(3) Equation (3) is an equation with separable variables. After we found its particular solution v = v(x), substitute it into (2). Since it satisfies equation (3), the expression in parentheses becomes zero. We get: 
This is also a separable equation. We find its general solution, and with it the solution to the original equation y = uv.64. Equation in total differentials. Integrating factor. Solution methods
65. Ordinary differential linear equations of higher orders: homogeneous and inhomogeneous. Linear differential operator, its properties (with proof).
Substituting this equation into x=C(y)e^(\sin(y)) , we obtain a general solution to the original equation, and therefore to this equation: Bernoulli's equation
From here we obtain the general integral of this equation
